1. Suppose y varies directly with x. Find each constant of variation.
Inverse Variation
ALGEBRA 1 LESSON 12-1
(For help, go to Lesson 5-5.)
Suppose y varies directly with x. Find each constant of variation.
1. y = 5x 2. y = –7x 3. 3y = x y = x
Write an equation of the direct variation that includes the given point.
5. (2, 4) 6. (3, 1.5) 7. (–4, 1) 8. (–5, –2)
12-1

2. 1. y = 5x; constant of variation = 5
Inverse Variation
ALGEBRA 1 LESSON 12-1
1. y = 5x; constant of variation = 5
2. y = –7x; constant of variation = –7
y = x
• 3y = • x
y = x; constant of variation =
y = x
y = x
4 • y = 4 • x
y = 4x; constant of variation = 4
Solutions 1 3 1 3 1 3 1 3 1 4 1 4
12-1

3. Solutions (continued)
Inverse Variation
ALGEBRA 1 LESSON 12-1
Solutions (continued)
5. Point (2, 4) in y = kx: 4 = k(2), so k = 2 and y = 2x.
6. Point (3, 1.5) in y = kx: 1.5 = k(3), so k = 0.5 and y = 0.5x.
7. Point (–4, 1) in y = kx: 1 = k(–4), so k = – and y = – x.
8. Point (–5, –2) in y = kx: –2 = k(–5), so k = and y = x. 1 4 1 4 2 5 2 5
12-1

4. xy = k Use the general form for an inverse variation.
ALGEBRA 1 LESSON 12-1
Suppose y varies inversely with x, and y = 9 when x = 85. Write an equation for the inverse variation.
xy = k Use the general form for an inverse variation.
(8)(9) = k Substitute 8 for x and 9 for y.
72 = k Multiply to solve for k.
xy = 72 Write an equation. Substitute 72 for k in xy = k.
The equation of the inverse variation is xy = 72 or y = . 72 x
12-1

5. 5(6) = 3y2 Substitute 5 for x1, 6 for y1, and 3 for x2.
Inverse Variation
ALGEBRA 1 LESSON 12-1
The points (5, 6) and (3, y) are two points on the graph of an inverse variation. Find the missing value.
x1 • y1 = x2 • y2 Use the equation x1 • y1 = x2 • y2 since you know coordinates, but not the constant of variation.
5(6) = 3y2 Substitute 5 for x1, 6 for y1, and 3 for x2.
30 = 3y2 Simplify.
10 = y2 Solve for y2.
The missing value is 10. The point (3, 10) is on the graph of the inverse variation that includes the point (5, 6).
12-1

6. Inverse Variation
ALGEBRA 1 LESSON 12-1
Jeff weighs 130 pounds and is 5 ft from the lever’s fulcrum. If Tracy weighs 93 pounds, how far from the fulcrum should she sit in order to balance the lever?
Relate:  A weight of 130 lb is 5 ft from the fulcrum. A weight of 93 lb is x ft from the fulcrum. Weight and distance vary inversely.
Define:  Let weight1 = 130 lb Let weight2 = 93 lb Let distance1 = 5 ft Let distance2 = x ft
12-1

7. Write: weight1 • distance1 = weight2 • distance2
Inverse Variation
ALGEBRA 1 LESSON 12-1
(continued)
Write: weight1 • distance1 = weight2 • distance2
130 • 5 = 93 • x Substitute.
650 = 93x Simplify.
6.99 = x Simplify.
= x Solve for x.
650 93
Tracy should sit 6.99, or 7 ft, from the fulcrum to balance the lever.
12-1

8. The values of y seem to vary inversely with the values of x. x y 3 10
Inverse Variation
ALGEBRA 1 LESSON 12-1
Decide if each data set represents a direct variation or an inverse variation. Then write an equation to model the data.
The values of y seem to vary inversely with the values of x.
x y
3 10
5 6
10 3 a.
Check each product xy.
xy: 3(10) = 30    5(6) = 30    10(3) = 30
The product of xy is the same for all pairs of data.
So, this is an inverse variation, and k = 30.
The equation is xy = 30.
12-1

9. The values of y seem to vary directly with the values of x.
Inverse Variation
ALGEBRA 1 LESSON 12-1
(continued)
x y
2 3
4 6
8 12 b.
The values of y seem to vary directly with the values of x.
Check each ratio . y x 6 4
= 1.5 12 8 y x 3 2
The ratio is the same for all pairs of data. y x
So, this is a direct variation, and k = 1.5.
The equation is y = 1.5x.
12-1

10. a. You buy several souvenirs for $10 each.
Inverse Variation
ALGEBRA 1 LESSON 12-1
Explain whether each situation represents a direct variation or an inverse variation.
a. You buy several souvenirs for $10 each.
The cost per souvenir times the number of souvenirs equals the total cost of the souvenirs.
Since the ratio is constant at $10 each, this is a direct variation.
cost
souvenirs
b. The cost of a $25 birthday present is split among several friends.
The cost per person times the number of people equals the total cost of the gift.
Since the total cost is a constant product of $25, this is an inverse variation.
12-1

11. 25. inverse variation; xy = 60 26. inverse variation; xy = 72
ALGEBRA 1 LESSON 12-1
pages 640–642  Exercises
1. xy = 18
2. xy = 2
3. xy = 56
4. xy = 1.5
5. xy = 24
6. xy = 7.7
7. xy = 2
8. xy = 0.5
9. xy = 0.06
10. 8
12. 6
13. 7
14. 3
16. 12
17. 96
19. 2
20.
21. 20
22. 3 h
mi/h
24. direct variation; y = 0.5x
25. inverse variation; xy = 60
26. inverse variation; xy = 72
27. Direct variation; the ratio
is constant at $1.79.
28. Inverse variation; the total
number of slices is constant at 8.
29. Inverse variation; the product
of the length and width remains
constant with an area of
24 square units.
30. 32; xy = 32
; rt = 1.1
; xy = 2.5
cost
pound 1 6
12-1

12. 41. direct variation; y = 0.4x; 8 42. direct variation; y = 70x; 0.9
Inverse Variation
ALGEBRA 1 LESSON 12-1
41. direct variation; y = 0.4x; 8
42. direct variation; y = 70x; 0.9
43. inverse variation; xy = 48; 0.5
44. a. greater
b. greater
c. less
45. a. 16 h; 10 h; 8 h; 4 h
b. hr worked, rate of pay
c. rt = 80
46. Check students’ work. L
48. p: y = 0.5x; q: xy = 8
49. a. y is doubled.
b. y is halved.
33. 1; ab = 1
; pq = 15.6
; xy = 375
36. Direct variation; the ratio
of the perimeter to the side
length is constant at 3.
37. Inverse variation; the product
of the rate and the time
is always 150.
38. Direct variation; the ratio of
the circumference to the
radius is constant at 2 . ft
days
12-1

13. 54. [2] Direct variation: y = kx, 10 = 5k, k = 2. So when x = 8,
Inverse Variation
ALGEBRA 1 LESSON 12-1
50. 4; s d = sd2 = k, so s =
51. a. x4y = k
b = k
52. C
53. F
54. [2] Direct variation: y = kx,
10 = 5k, k = 2. So when x = 8,
y = 2 • 8 = 16. Inverse variation:
xy = k, 5 • 10 = 50,
So when x = 8, y = , or 6.25.
[1] no work shown OR
one computational error 1 2 2 1 4 k d2
55. [4] a.
b. The variables speed and
time are inversely related.
c h
[3] one computational error
[2] one part missing
[1] two parts missing
56.
57.
x4y z 1 2 50 8 8 17 15 17
12-1

14. Inverse Variation
ALGEBRA 1 LESSON 12-1
58.
59.
60.
61. 15 17
68. (3a – 1)(a + 4)
69. (5x + 2)(3x + 7)
70. (2y – 3)(y + 8) 8 17 8 15 15 8
12-1

15. 3. Write an equation to model the data and complete the table.
Inverse Variation
ALGEBRA 1 LESSON 12-1
1. The points (5, 1) and (10, y) are on the graph of an inverse variation. Find y.
2. Find the constant of variation k for the inverse variation where a = 2.5 when b = 7.
0.5
17.5
x y 1 2 6 3 9
3. Write an equation to model the data and complete the table.
xy = 1 3 18
4. Tell whether each situation represents a direct variation or an inverse variation.
a. You buy several notebooks for $3 each.
b. The $45 cost of a dinner at a restaurant is split among several people.
direct variation
Inverse variation
12-1

16. Graphing Rational Functions
ALGEBRA 1 LESSON 12-2
(For help, go to Lessons 5-3, 8-7, and 10-1.)
Evaluate each function for x = –2, 0, 3.
1. ƒ(x) = x – 8 2. g(x) = x y = 3x
Graph each function.
4. ƒ(x) = 2x g(x) = –x2 6. y = 2x
12-2

17. Graphing Rational Functions
ALGEBRA 1 LESSON 12-2
Solutions
1. ƒ(x) = x – 8 for x = –2, 0, 3:
ƒ(–2) = –2 – 8 = –10
ƒ(0) = 0 – 8 = –8
ƒ(3) = 3 – 8 = –5
2. g(x) = x2 + 4 for x = –2, 0, 3:
g(–2) = (–2)2 + 4 = = 8
g(0) = = = 4
g(3) = = = 13
3. y = 3x for x = –2, 0, 3:
y = 3–2 = = =
y = 30 = 1
y = 33 = 3 • 3 • 3 = 9 • 3 = 27 1 32 1
3 • 3 1 9
12-2

18. Graphing Rational Functions
ALGEBRA 1 LESSON 12-2
Solutions (continued)
4. ƒ(x) = 2x + 1
5. g(x) = – x2
6. y = 2x
12-2

19. Graphing Rational Functions
ALGEBRA 1 LESSON 12-2
The function t = models the time it will take you to travel 70 miles at different rates of speed. Graph this function. 70 r
r t
Step 1: Make a table of values.
Step 2: Plot the points.
12-2

20. Graphing Rational Functions
ALGEBRA 1 LESSON 12-2 4
x – 3
Identify the vertical asymptote of y = Then graph the function.
Step 1: Find the vertical asymptote.
x – 3 = 0 The numerator and denominator have no common factors. Find any value(s) where the denominator equals zero.
x = 3 This is the equation of the vertical asymptote.
12-2

21. Graphing Rational Functions
ALGEBRA 1 LESSON 12-2
(continued)
Step 2:  Make a table of values. Use values of x near 3, the asymptote.
Step 3: Graph the function. 4 3 –
x y
2 –4
1 –2
–1 –1
4 4
5 2 6
12-2

22. Graphing Rational Functions
ALGEBRA 1 LESSON 12-2 4
x + 4
Identify the asymptotes of y = Then graph the function.
Step 1: From the form of the function, you can see that there is a vertical asymptote at x = –4 and a horizontal asymptote at y = 3. Sketch the asymptotes.
Step 2:  Make a table of values using values of x near –4. 1 3
x y
–10 2
–8 2
–6 1
–5 –1
–3 7
–2 5
–1 4
2 3 2
12-2

23. Graphing Rational Functions
ALGEBRA 1 LESSON 12-2
(continued)
Step 3: Graph the function.
12-2

24. Graphing Rational Functions
ALGEBRA 1 LESSON 12-2
Describe the graph of each function.
a. y = x 6
The graph is a line with slope and y-intercept 0. 1 6
b. y = 6x
The graph is of exponential growth.
c. y = 6x2
The graph is a parabola with axis of symmetry at x = 0.
d. y = | x – 6|
The graph is an absolute value function with a vertex at (6, 0).
12-2

25. Graphing Rational Functions
ALGEBRA 1 LESSON 12-2
(continued)
e. y = 6 x
The graph is a rational function with vertical asymptote at x = 0 and horizontal asymptote at y = 0.
f. y = x – 6
The graph is the radical function y = x shifted right 6 units.
g. y = 6x
The graph is a line with slope 6 and y-intercept 0.
12-2

26. Graphing Rational Functions
ALGEBRA 1 LESSON 12-2
pages 648–650  Exercises 1. 2. 5.
6. 0
7. 2
8. –2
9. 2
10. x = 2, y = 0
11. x = –1, y = 0
12. x = 1, y = –1
13. x = 0, y = 2 3. 4.
12-2

27. Graphing Rational Functions
ALGEBRA 1 LESSON 12-2
14. x = 0;
15. x = 0;
16. x = –1;
17. x = 5;
18. x = –4;
19. x = –4;
12-2

28. Graphing Rational Functions
ALGEBRA 1 LESSON 12-2
20. x = 0, y = –5;
21. x = 0, y = 5;
22. x = 0, y = –6;
23. x = –1, y = 4;
24. x = 3, y = –5;
25. x = 1, y = –2;
12-2

29. Graphing Rational Functions
ALGEBRA 1 LESSON 12-2
26. line with slope 4, y-int. 1
27. absolute value function with vertex (4, 0)
28. exponential decay
29. line with slope , y-int. 0
30. rational function, with asymptotes
x = 0, y = 1
31. radical function; y = x
shifted right 4, up 1
32. parabola with axis of
symmetry x = 0
33. rational function with
asymptotes x = –4, y = –1
34. parabola with axis of symmetry x = –
35. moves graph 1 unit to the left 1 4
36. moves graph 3 units to the right
37. lowers graph 15 units
38. moves graph 12 units left
39. moves the graph up 12 units
40. moves the graph left 3 units
41. moves the graph down 2 units
42. moves the graph 3 units left
and 2 units down
12-2

30. Graphing Rational Functions
ALGEBRA 1 LESSON 12-2
43. x = 0, y = 0;
44. x = 0, y = 0;
45. x = –4, y = 0;
46. x = 0, y = 1;
47. x = –1, y = 4;
48. x = –1, y = –3;
12-2

31. Graphing Rational Functions
ALGEBRA 1 LESSON 12-2
49. x = 1, y = 3;
50. x = –5, y = 1;
51. x = 3, y = –2;
52. Answers may vary. Sample: ƒ(x) = , g(x) =
lumens; 1.97 lumens
54. a.
b. x = 0, y = 0; x = 0, y = 0
c. y is any real number except 0; y > 0. 1 x
12-2

32. Graphing Rational Functions
ALGEBRA 1 LESSON 12-2
55. a.
d 40
b. 16; 1600; 160,000
c. The signal is extremely strong
when you are in the immediate
vicinity of a transmitter and it will
interfere with the other station.
56. The graph of y = and y = – are both composed of
two curves with asymptotes x = 0 and y = 0. The graph of
y = – is a reflection of the graph of y = over the y-axis.
57.
58.
> – 3 x 3 x 3 x 3 x
12-2

33. Graphing Rational Functions
ALGEBRA 1 LESSON 12-2
59.
60.
61. a. x = –3, y = –2
b. y = – 2
62.
No; ƒ(x) = is equivalent to
g(x) = x + 1 for all values except x = –2.
63. C
64. I
65. A
66. C
(x + 2)(x + 1)
x + 2 1
x + 3
12-2

34. Graphing Rational Functions
ALGEBRA 1 LESSON 12-2
67. [2] a. The graph of g(x) =
is a translation of ƒ(x) = units up and 1 unit right.
b. x = 1 and y = 5
[1] one part answered correctly
68. xy = 21
69. xy = 16
70. xy = 22
71. xy = 21.08
72. 0
73. 1
74. 2 4
x – 1
75. 3(d – 6)(d + 6)
76. 2(m – 12)(m + 5)
77. (t 2 + 3)(t – 1) 4 x
12-2

35. Graphing Rational Functions
ALGEBRA 1 LESSON 12-2
3. Describe the graph of each function.
1. Identify the vertical asymptote of y = Then graph the function. 3
x + 2
2. Identify the vertical and horizontal asymptotes of y = – 3. Then graph the function. 1
x – 2
a. y = x + 3
b. y =
c. y =
d. y = 3x2 x
12-2

36. Graphing Rational Functions
ALGEBRA 1 LESSON 12-2
1. Identify the vertical asymptote of y = Then graph the function. 3
x + 2
x = –2
2. Identify the vertical and horizontal asymptotes of y = – 3. Then graph the function. 1
x – 2
x = 2; y = –3
12-2

37. Graphing Rational Functions
ALGEBRA 1 LESSON 12-2
3. Describe the graph of each function.
a. y = x + 3
b. y =
c. y =
d. y = 3x2
radical function y = x shifted left 3 units x 3
line with slope and y-intercept 0 1 3 3 x
rational function with vertical asymptote at x = 0 and horizontal asymptote at y = 2
parabola with axis of symmetry at x = 0.
12-2

38. Simplifying Rational Expressions
ALGEBRA 1 LESSON 12-3
(For help, go to Lessons 9-5 and 9-6.)
Write each fraction in simplest form. 8 2 15 24 – 25 35 1. 2. 3.
Factor each quadratic expression.
4. x2 + x – x2 + 6x x2 – 2x – 15
7. x2 + 8x x2 – x – x2 – 7x + 12
12-3

39. Simplifying Rational Expressions
ALGEBRA 1 LESSON 12-3
Solutions 8 2 15 24 –
3 • 5
3 • 8 5 8
1. = 8 ÷ 2 = 4
2. = – = – 25 35
3. = =
5 • 5
5 • 7 5 7
4. Factors of –12 with a sum of 1: 4 and –3. x2 + x – 12 = (x + 4)(x – 3)
5. Factors of 8 with a sum of 6: 2 and 4. x2 + 6x + 8 = (x + 2)(x + 4)
6. Factors of –15 with a sum of –2: 3 and –5. x2 – 2x – 15 = (x + 3)(x – 5)
12-3

40. Simplifying Rational Expressions
ALGEBRA 1 LESSON 12-3
Solutions (continued)
7. Factors of 16 with a sum of 8: 4 and 4. x2 + 8x + 16 = (x + 4)(x + 4) or (x + 4)2
8. Factors of –12 with a sum of –1: 3 and –4. x2 – x – 12 = (x + 3)(x – 4)
9. Factors of 12 with a sum of –7: –3 and –4. x2 – 7x + 12 = (x – 3)(x – 4)
12-3

41. Simplifying Rational Expressions
ALGEBRA 1 LESSON 12-3
3x + 9
x + 3
Simplify
Factor the numerator. The denominator cannot be factored.
3x + 9
x + 3
3(x + 3) =
Divide out the common factor x + 3.
3(x + 3)
x + 3 1 =
= 3 Simplify.
12-3

42. Simplifying Rational Expressions
ALGEBRA 1 LESSON 12-3
Simplify
4x – 20
x2 – 9x + 20
4x – 20
x2 – 9x + 20
4(x – 5)
(x – 4) (x – 5) =
Factor the numerator and the denominator.
Divide out the common factor x – 5.
4(x – 5)
(x – 4) (x – 5) 1 = 4
x – 4 =
Simplify.
12-3

43. Simplifying Rational Expressions
ALGEBRA 1 LESSON 12-3
Simplify
3x – 27
81 – x2
3x – 27
81 – x2
3(x – 9)
(9 – x) (9 + x) =
Factor the numerator and the denominator.
3(x – 9)
– 1 (x – 9) (9 + x)
Factor –1 from 9 – x. =
3(x – 9)
– 1 (x – 9) (9 + x) 1
Divide out the common factor x – 9. = 3
9 + x
= –
Simplify.
12-3

44. Simplifying Rational Expressions
ALGEBRA 1 LESSON 12-3
The baking time for bread depends, in part, on its size and shape. A good approximation for the baking time, in minutes, of a cylindrical loaf is , or , where the radius r and the length h of the baked loaf are in inches. Find the baking time for a loaf that is 8 inches long and has a radius of 3 inches. Round your answer to the nearest minute.
60 • volume
surface area
30rh
r + h
30rh
r + h
30 (3) (8)
3 + 8 =
Substitute 8 for r and 3 for h.
Simplify.
720 11 =
Round to the nearest whole number. 65
The baking time is approximately 65 minutes.
12-3

45. Simplifying Rational Expressions
ALGEBRA 1 LESSON 12-3
pages 654–656  Exercises 1. 2. 3. 4.
5. 3x 6. 7. 8. 9.
2a + 3 4 1 7x 3 2
x + 2 x2
b + 4
m – 7
10.
11.
12.
13.
14. b + 3
15.
16. –1
17.
18. –2
19. – w
w – 7
a + 1 5
m + 3
m + 2
c – 4
c + 3
m – 2 –4
t + 1
20. –
21. –
min
min
min
25.
26.
27.
28.
29.
w – 4
2r – 1
r + 5
v + 5
7z + 2
z – 1
4a2
2a – 1
5t – 4
3t – 1
3(z + 4) z3
12-3

46. Simplifying Rational Expressions
ALGEBRA 1 LESSON 12-3
30.
31. –
32.
33.
34. Answers may vary. Sample:
35. a. i.  ii. 
b. ;
36. The student canceled terms
instead of factors.
37. –3 is not in the domain of
38.
39.
40.
41.
42.
43.
44.
45. sometimes
46. sometimes
47. never
48. C
49. I
50. B
51. D
52. C
53. [2] The student put the
4 in the numerator
rather than in
the denominator. =
[1] no explanation OR
incorrectly simplified
expression
2s + 1 s2 5w
5w + 6
2a + 1
a + 3 1 4 3y
4(y + 4)
4 + 3m
m – 7
–c(3c + 5)
5c + 4
t + 3
3(t + 2)
m – n
m + 10n 3
(x – 2)(x + 3)
a – 3b
a + 4b
2b + 4h bh
2h + 2r rh
6v – 7w
3v – 2w
x – 5
4x – 20 4 9 4 9
x – 5
4(x – 5) 1 4
x2 – 9
x + 3
12-3

47. Simplifying Rational Expressions
ALGEBRA 1 LESSON 12-3
54. vertical asymptote: x = 0;
horizontal asymptote: y = 2;
55. vertical asymptote: x = 4;
horizontal asymptote: y = 0;
56. vertical asymptote: x = 0;
horizontal asymptote: y = –4;
58. a2b3c4 b
60.
61. y = x2, y = –2x2, y = 3x2
62. y = x2, y = x2, y = x2
63. y = 0.5x2, y = 2x2, y = –4x2
64. y = –x2, y = 2.3x2, y = –3.8x2 2
5m2 1 4 3 5
12-3

48. Simplifying Rational Expressions
ALGEBRA 1 LESSON 12-3
Simplify each expression. 1. 3. 2.
6 – 2x
x – 3
x2 + 8x
x2 – 64
x – 6
36 – x2 x
x – 8 –1
x + 6 –2 4. 5.
6x2 – x – 12
8x2 – 10x – 3
4x2 + x x3
3x + 4
4x + 1
4x + 1 x2
12-3

49. Multiplying and Dividing Rational Expressions
ALGEBRA 1 LESSON 12-4
(For help, go to Lessons 8-3 and 9-6.)
Simplify each expression.
1. r 2 • r b3 • b4 3. c7 ÷ c2
4. 3x4 • 2x5 5. 5n2 • n a3 (–3a2)
Factor each polynomial.
7. 2c2 + 15c t2 – 26t q2 + 11q + 5
12-4

50. Multiplying and Dividing Rational Expressions
ALGEBRA 1 LESSON 12-4
Solutions
1. r 2 • r 8 – r (2 + 8) = r b3 • b4 = b(3+4) = b7
3. c7 ÷ c2 = c(7 – 2) = c x4 • 2x5 = (3 • 2)(x4 • x5) = 6x(4 + 5) = 6x9
5. 5n2 • n2 = 5(n2 • n2) = 5n(2 + 2) = 5n4
6. 15a3(–3a2) = 15(–3)(a3 • a2) = –45a(3 + 2) = –45a5
7. 2c2 + 15c + 7 = (2c + 1)(c + 7) Check: (2c + 1)(c + 7) = 2c2 + 14c + 1c + 7 = 2c2 + 15c + 7
8. 15t2 – 26t + 11 = (15t – 11)(t – 1) Check: (15t – 11)(t – 1) = 15t2 – 15t – 11t = 15t2 – 26t + 11
9. 2q2 + 11q + 5 = (2q + 1)(q + 5) Check: (2q + 1)(q + 5) = 2q2 + 10q + 1q + 5 = 2q2 + 11q + 5
12-4

51. Multiplying and Dividing Rational Expressions
ALGEBRA 1 LESSON 12-4
Multiply.
a • 7 y 8 y2 7 y 8 y2 • = 56 y3
Multiply the numerators and multiply the denominators.
b • x
x + 5
x – 2
x – 6
Multiply the numerators and multiply the denominators. Leave the answer in factored form. x
x + 5
x – 2
x – 6
x(x – 2)
(x + 5) (x – 6) = •
12-4

52. Multiplying and Dividing Rational Expressions
ALGEBRA 1 LESSON 12-4
3x +1 4 8x
9x2 – 1
Multiply and
3x + 1 4 8x
9x2 – 1 • =
(3x – 1) (3x + 1)
Factor denominator.
Divide out the common factors (3x +1) and 4. = 8x
(3x – 1) (3x + 1) 1 2
3x +1 4 • =
Simplify. 2x
3x – 1
12-4

53. Multiplying and Dividing Rational Expressions
ALGEBRA 1 LESSON 12-4
5x + 1
3x + 12
Multiply and x2 + 7x + 12.
5x + 1
3x + 12
• (x2 + 7x + 12) =
3 (x + 4) •
(x + 3) (x + 4) 1
Factor.
Divide out the common factor x + 4.
5x + 1
3 (x + 4) = •
(x + 3) (x + 4) 1
(5x +1) (x + 3) 3 =
Leave in factored form.
12-4

54. Multiplying and Dividing Rational Expressions
ALGEBRA 1 LESSON 12-4
x2 + 13x +40
x – 7
x + 8
x2 – 49
Divide by
x2 – 49
x + 8 ÷ = •
Multiply by , the reciprocal of
x2 + 13x +40
x – 7 =
(x + 5) (x + 8)
x – 7
(x + 7) (x – 7)
x + 8 •
Factor.
Divide out the common factors x + 8 and x – 7. =
(x + 8) (x + 5)
x – 7
(x + 7) (x – 7)
x + 8 • 1
Leave in factored form.
= (x + 5) (x + 7)
12-4

55. Multiplying and Dividing Rational Expressions
ALGEBRA 1 LESSON 12-4
Divide by (8x2 + 16x).
x2 + 9x + 14
11x
x2 + 9x + 14
11x
8x2 + 16x 1 ÷ = •
Multiply by the reciprocal of 8x2 + 16x
Factor. 1
8x (x + 2)
(x + 7) (x + 2)
11x • =
Divide out the common factor x + 2. = 1
8x (x + 2)
(x + 7) (x + 2)
11x • =
x + 7
88x2
Simplify.
12-4

56. Multiplying and Dividing Rational Expressions
ALGEBRA 1 LESSON 12-4
pages 659–661  Exercises 1. 2. 3. 4. 5. 6. 7. 8. 9.
10.
11.
12.
13. 4(t + 1)(t + 2)
14. 3(2m + 1)(m + 2)
15.
16.
17. –
18.
19.
20.
21. 6
22. –
23. –
24.
25.
26.
27.
28.
29. t + 3 1 3
x – 1
x + 3
35x 36 1 2 1 2 12
t 2
3(4x + 1)
x – 1 1 3 40
3a5
m(m – 2)
(m + 2)(m – 1)
2(x + 2)
x – 1
(x – 1)(x – 2) 3
n – 3
4n + 5
2x(x – 1)
3(x + 1)
x + 1 2 3 x
12x2
5(x + 1)
2d – 5
6d 2 11
7k – 15 2c
c – 1 1
c2 – 1 1
x + 1
5x4 2 1
s + 4 9 t
12-4

57. Multiplying and Dividing Rational Expressions
ALGEBRA 1 LESSON 12-4
30.
31.
32.
33.
34.
35. The student forgot to rewrite
the expression using the
reciprocal before canceling.
36. Answers may vary.
Sample: • ;
37. 0, 4, –4
38. $88.71
c + 1
c – 1
39. $132.96
40. a. $100,000
b. 360 payments
c. $599.55
d. $215,838
41.
42.
43.
44.
45. She wrote w5 as a fraction so
she could easily see what
she could cancel.
3t – 5
7t 2
5(2x – 5)
x – 5
x – 2
x – 3
x – 2
4(x + 7)
x – 5 x
2m2(m + 2)
(m – 1)(m + 4) 2
a + 5
r + 3
(r – 1)(r + 1)2
9(m + 1)
3m2
m + 2
3(m + 1)
m + 2 m2
12-4

58. Multiplying and Dividing Rational Expressions
ALGEBRA 1 LESSON 12-4
46. a. b.
47.
48. 1
49.
50.
51.
52.
53.
54. B
55. G
x(3x + 2)
4(2x + 1)2
x(x – 2)
2(x – 1)
m – 2
2m(m – 1)
x 2 1
(w + 2)(w + 3)
–(2a + 3b)(a + 2b)
(5a + b)(2a – 3b)
9m 2(m + 1) 2
56. D
57. G
58. [2] • =
= 3(x + 1)
[1] one computational error
OR answer with no work shown
59.
60.
61.
62.
63.
x 2 – 1 x 3x
x – 1
(x + 1)(x – 1)(3x)
x(x – 1)
b – 5 3 4k 7 q4 4
5t 2 – 9 8
y + 5
12-4

59. Multiplying and Dividing Rational Expressions
ALGEBRA 1 LESSON 12-4
64.
65.
66.
67.
70. 5
71. 11 1
2a2 – 3
2z + 3
z + 1
2c – 9
2c + 8
76.
74.
75. m2
2 – 3m
12-4

60. Multiplying and Dividing Rational Expressions
ALGEBRA 1 LESSON 12-4
Multiply or divide. 1. 2.
7x 2 5 15
14x • 3x 2
6x + 3
x + 6 •
x2 + 9x + 18
2x + 1
3(x + 3) 3. 4.
x + 3
x + 1 ÷
(x2 + 5x + 6) 1
(x + 1)(x + 2)
4x + 8 3x •
9x 2
x + 2
12x 5.
2x + 4
x2 + 11x + 18 ÷
x + 1
x2 + 14x + 45
2(x + 5)
x + 1 6.
(x2 + 12x + 11) •
x + 9
x2 + 20x + 99
x + 1
12-4

61. Write each polynomial in standard form. 1. 9a – 4a2 + 1
Dividing Polynomials
ALGEBRA 1 LESSON 12-5
(For help, go to Lessons 9-1 and 9-3.)
Write each polynomial in standard form.
1. 9a – 4a2 + 1
2. 3x2 – 6 + 5x – x3
3. –2 + 8t
Find each product.
4. (2x + 4)(x + 3)
5. (–3n – 4)(n – 5)
6. (3a2 + 1)(2a – 7)
12-5

62. Dividing Polynomials Solutions 1. 9a – 4a2 + 1 = –4a2 + 9a + 1
ALGEBRA 1 LESSON 12-5
Solutions
1. 9a – 4a2 + 1 = –4a2 + 9a + 1
2. 3x2 – 6 + 5x – x3 = –x3 + 3x2 + 5x – 6
3. –2 + 8t = 8t – 2
4. (2x + 4)(x + 3) = (2x)(x) + (2x)(3) + (4)(x) + (4)(3) = 2x2 + 6x + 4x + 12 = 2x2 + 10x + 12
5. (–3n – 4)(n – 5) = (–3n)(n) + (–3n)(–5) + (–4)(n) + (–4)(–5) = –3n2 + 15n – 4n + 20 = –3n2 + 11n + 20
6. (3a2 + 1)(2a – 7) = (3a2)(2a) + (3a2)(–7) + (1)(2a) + (1)(–7) = 6a3 – 21a2 + 2a – 7
12-5

63. Multiply by the reciprocal of 3x2.
Dividing Polynomials
ALGEBRA 1 LESSON 12-5
Divide (18x3 + 9x2 – 15x) by 3x2.
(18x3 + 9x2 – 15x)
3x2 ÷ =
(18x3 + 9x2 – 15x) • 1
Multiply by the reciprocal of 3x2.
= + –
18x3
3x2
9x2
15x
Use the Distributive Property.
= 6x1 + 3x0 – 5 x
Use the division rules for exponents.
= 6x + 3 – 5 x
Simplify.
12-5

64. Step 1: Begin the long division process.
Dividing Polynomials
ALGEBRA 1 LESSON 12-5
Divide (5x2 + 2x – 3) by (x + 2)
x x2 + 2x – 3
Divide: Think 5x2 ÷ x = 5x.
Step 1: Begin the long division process. 5x
Align terms by their degree. So put 5x above 2x of the dividend.
5x2 + 10x
Multiply: 5x(x + 2) = 5x2 + 10x. Then subtract.
– 8x – 3
Bring down – 3.
12-5

65. Step 2: Repeat the process: Divide, multiply, subtract, bring down.
Dividing Polynomials
ALGEBRA 1 LESSON 12-5
(continued)
Step 2: Repeat the process: Divide, multiply, subtract, bring down.
5x – 8
x x2 + 2x – 3
5x2 + 10x
Divide: –8x ÷ x = – 8
– 8x – 3
Multiply: – 8(x + 2) = – 8x – 16. Then subtract.
– 8x – 16 13
The remainder is 13.
The answer is 5x – 13
x + 2
12-5

66. Since A = w, divide the area by the width to find the length.
Dividing Polynomials
ALGEBRA 1 LESSON 12-5
The width and area of a rectangle are shown in the figure below. What is the length?
Since A = w, divide the area by the width to find the length.
6x3 – 9x2
4x2 + 0x
3x2
4x2 + 6x
–6x – 9
+ 2x
+ 3
–6x – 9
2x – 3 6x3 – 5x2 + 0x – 9 Rewrite the dividend with 0x.
The length of the rectangle is (3x2 + 2x + 3) in.
12-5

67. Divide (–8x – 2 + 6x2) by (–1 + x).
Dividing Polynomials
ALGEBRA 1 LESSON 12-5
Divide (–8x – 2 + 6x2) by (–1 + x).
Rewrite –8x – 2 + 6x2 as 6x2 – 8x – 2 and –1 + x as x – 1.
Then divide. 6x
x – 1
6x2 – 8x – 2
6x2 – 6x
– 2
–2x –2
–2x + 2 –4
The answer is 6x – 2 – 4
x – 1
12-5

68. Dividing Polynomials pages 664–666 Exercises 11. 3x – 1
ALGEBRA 1 LESSON 12-5
pages 664–666  Exercises
1. x4 – x3 + x2
2. 3x4 –
3. 3c2 + 2c –
4. n2 – 18n + 3
5. 4 –
6. –t 3 + 2t 2 – 4t + 5
7. x – 3
8. 2t + 9 +
9. n – 1
10. y – 3 +
11. 3x – 1
12. –2q – 10 +
13. 5t – 50
14. 2w2 + 2w + 5 –
15. b2 – 3b – 1 +
16. c2 –
17. t 2 – 2t – 2
18. n2 – 2n – 21 –
19. (r 2 + 5r + 1) cm
20. (4c2 – 8c + 16) ft
21. b 22
2q + 1 2 x 1 3 10
w – 1 3
3b – 1 16 q 1
c – 1 8
n + 2 16
t – 3 8
y + 2 1
b + 4
12-5

69. 41. a. Answers may vary. Sample: (c3 + 3c2 – 2c – 4); (c + 1)
Dividing Polynomials
ALGEBRA 1 LESSON 12-5
22. a – 1 –
23. 10w – 681 +
24. t –
25. 2x2 + 5x + 2
26. 3q 2 + 2q + 3 +
27. 3x + 2 –
28. c2 + 11c – 15 +
29. 2b2 + 2b
30. y2 + 5y
31. 28a – 12
32. 5t 3 – 25t t – 575 + 1 2x 8 c 12
q – 2 2
a + 4
49,046
w + 72 10
b – 1
138
y – 5 9
t + 4
2881
t + 5
33. k 2 – 0.3k – 0.4
34. 3s – 8 +
35. –2z2 + 3z – 4 +
36. 6m2 – 24m + 99 –
37. –16c2 – 20c – 25
38. 2r 4 + r 2 – 7
39. t – 1 +
40. z3 – 3z2 + 10z – 30 +
41. a. Answers may vary. Sample:
(c3 + 3c2 – 2c – 4); (c + 1)
b. (c3 + 3c2 – 2c – 4) ÷ (c + 1)
= c2 + 2c – 4 29
2s + 3 5
z + 1
326
m + 4 2t
2t 3 + 1 88
z + 3
12-5

70. b. Answers may vary. Sample: x –2 –1 0 1 2 y 1
Dividing Polynomials
ALGEBRA 1 LESSON 12-5
42. a. y = 2 –
b. Answers may vary. Sample:
x –2 –
y 1
c. vertical asymptote: x = –3
horizontal asymptote: y = 2
43. The binomial is a factor of the
polynomial if there is no
remainder from the division. 1
x + 3
44. m2 + 5m + 4
45. a. d – 2 +
b. d 2 – 2d + 3 –
c. d 3 – 2d 2 + 3d – 4 +
d. Answers may vary. Sample:
d 4 – 2d 3 + 3d 2 – 4d + 5 –
e. d 4 – 2d 3 + 3d 2 – 4d + 5 –
46. 12
47. a. t =
b. t 2 – 7t + 12
48. 2a2b2 – 3ab3 + 5ab2
49. 3x + 2y 3 2 5 7 4 9
d + 1 6 d r
12-5

71. [1] one computational error OR correct answer with no work shown x + 1
Dividing Polynomials
ALGEBRA 1 LESSON 12-5
50. 10r 5 + 2r 4 + 5r 2
51. 2b3 – 2b2 + 3
52. a. x – 3 +
b. ƒ(x) = x – 3 +
c. y = x – 3 d.
53. C
54. G
55. B
56. [2] x2 + 3x + 2
2x – 1 2x3 + 5x2 + x – 2
2x3 – x2
6x2 + x
6x2 – 3x
4x – 2
x + 2 x2 + 3x + 2
x2 + 2x
x + 2
The width is x + 1.
[1] one computational error OR
correct answer with no work shown 10
x + 5
x + 1
12-5

72. b. Tables may vary. Sample:
Dividing Polynomials
ALGEBRA 1 LESSON 12-5
57. [4] a
x + 4 3x + 10
3x + 12 –2
y = 3 –
b. Tables may vary. Sample:
vertical asymptote: x = –4
horizontal asymptote: y = 3;
[3] appropriate methods, but with one computational error
[2] no asymptotes OR no graph
[1] one part only
58. n + 2
59.
60.
61.
62. 5
64. 15
69. 17
70. –12.7
(t – 5)(3t + 1)(2t + 11)
(2t – 55)(t + 1)(3t)
3c + 8
2c + 7 2
x + 4
(x + 5)(x + 4)2
(x + 7)(x + 8)2
12-5

73. 1. (x8 – x6 + x4) ÷ x2 2. (4x2 – 2x – 6) ÷ (x + 1)
Dividing Polynomials
ALGEBRA 1 LESSON 12-5
Divide.
1. (x8 – x6 + x4) ÷ x2 2. (4x2 – 2x – 6) ÷ (x + 1)
3. (6x3 + 5x2 + 11) ÷ (2x + 3) 4. ( x3) ÷ (4x + 3)
x6 – x4 + x2
4x – 6
3x2 – 2x + 3 + 2
2x + 3
16x2 – 12x + 9 + 2
4x + 3
12-5

74. Adding and Subtracting Rational Expressions
ALGEBRA 1 LESSON 12-6
(For help, go to Lessons 1-5 and 906.)
Simplify each expression. 1. 4 9 2 + 2. 3 7 5 – 3. 1 2 5 2 + – 4. 5 6 2 9 + 5. 1 4 3 – 6. 12 4x 9 2x + 7. 7x 12 x – 8. 9. 7
12y 1
Factor each quadratic expression.
10. x2 + 3x + 2
11. y2 + 7y + 12
12. t 2 – 4t + 4
12-6

75. Adding and Subtracting Rational Expressions
ALGEBRA 1 LESSON 12-6
Solutions 1. 4 9 2 + =
4 + 2 6
3 • 2
3 • 3 3 2. 3 7 5 – =
3 – 5 –2 2 3. 1 2 + 5 – =
1 + (– 5) 7 –4
= –2 4. 5 6 2 9 +
5 • 3
6 • 3
2 • 2
9 • 2 15 18 = 19 or 4 1 5. 1 4 3 –
1 • 3
4 • 3
1 • 4
3 • 4 = 12
3 – 4 –1
= – 6. 5 12 3 4 5 12
3 • 3
4 • 3 5 12 9 12
5 – 9 12 –4 12 1 3 – = – = – = =
= –
12-6

76. Adding and Subtracting Rational Expressions
ALGEBRA 1 LESSON 12-6
Solutions (continued) 4x 9 2x + 7. =
4x + 2x 6x
3 • 2 • x
3 • 3 3 or 2 x 7x 12 x – 8. =
7x – x 6x 2 or 1
6 • x
6 • 2 9. 7
12y 1 – =
7 – 1
6 • 1
6 • 2y 6 2y
10. Factors of 2 with a sum of 3: 1 and 2. x2 + 3x + 2 = (x + 1)(x + 2)
11. Factors of 12 with a sum of 7: 3 and 4. y2 + 7y + 12 = (y + 3)(y + 4)
12. Factors of 4 with a sum of –4: –2 and –2. t2 – 4t + 4 = (t – 2)(t – 2) or (t – 2)2
12-6

77. Adding and Subtracting Rational Expressions
ALGEBRA 1 LESSON 12-6 4
x + 3 2
x + 3
Add and 4
x + 3 2 + =
4 + 2
Add the numerators. = 6
x + 3
Simplify the numerator.
12-6

78. Adding and Subtracting Rational Expressions
ALGEBRA 1 LESSON 12-6
3x + 5
3x2 + 2x – 8
4x + 7
3x2 + 2x – 8
Subtract from .
3x + 5
3x2 + 2x – 8
4x + 7 – =
4x + 7 – (3x + 5)
Subtract the numerators.
Use the Distributive Property.
4x + 7 – 3x + 5
3x2 + 2x – 8 =
Simplify the numerator.
x + 2
3x2 + 2x – 8 =
Factor the denominator. Divide out the common factor x + 2.
x + 2
(3x – 4) (x + 2) = 1
Simplify. 1
3x – 4 =
12-6

79. Adding and Subtracting Rational Expressions
ALGEBRA 1 LESSON 12-6 3 4x 1 8
Add + .
Step 1: Find the LCD of and 3 4x 1 8
4x = 2 • 2 • x Factor each denominator.
8 = 2 • 2 • 2
LCD = 2 • 2 • 2 • x = 8x
Step 2: Rewrite using the LCD and add.
Rewrite each fraction using the LCD. 3 4x 1 8 +
2 • 3
2 • 4x
1 • x
8 • x =
Simplify numerators and denominators. 6 8x x + =
Add the numerators. =
6 + x 8x
12-6

80. Adding and Subtracting Rational Expressions
ALGEBRA 1 LESSON 12-6 7
x + 4 3
x – 5
Add and .
Step 1:  Find the LCD of x + 4 and x – 5. Since there are no common factors, the LCD is (x + 4)(x – 5).
Step 2: Rewrite using the LCD and add. 7
x + 4 3
x – 5 + =
7 (x – 5)
(x + 4) (x – 5)
3 (x + 4)
Rewrite the fractions using the LCD.
Simplify each numerator.
7x – 35
(x + 4) (x – 5)
3x + 12 = +
Add the numerators.
7x – x + 12
(x + 4) (x – 5) =
Simplify the numerator. =
10x – 23
(x + 4) (x – 5)
12-6

81. Adding and Subtracting Rational Expressions
ALGEBRA 1 LESSON 12-6
The distance between Seattle, Washington, and Miami, Florida, is about 5415 miles. The ground speed for jet traffic from Seattle to Miami can be about 14% faster than the ground speed from Miami to Seattle. Use r for the jet’s ground speed. Write and simplify an expression for the round-trip air time.
Miami to Seattle time:
5415 r
time =
distance
rate
14% more than a number is 114% of the number.
Seattle to Miami time:
5415
1.14r
time =
distance
rate
12-6

82. Adding and Subtracting Rational Expressions
ALGEBRA 1 LESSON 12-6
(continued)
5415 r
5415
1.14r
An expression for the total time is +
5415 r
1.14r +
6173
Rewrite using the LCD, 1.14r.
Add the numerators. =
11588
1.14r
Simplify.
10165 r =
12-6

83. Adding and Subtracting Rational Expressions
ALGEBRA 1 LESSON 12-6
pages 669–671  Exercises 1. 2. 3. 4. 5. 6. 7.
8. – 9. 9 2m 7
6t – 1
n + 2
n + 3 14
c – 5
2s2 + 1
4s2 + 2
6c – 28
2c + 7 –3
2 – b 1
t 2 + 1
–2t
2t – 3
10. 1
11. 2
12. –1
13. 2x2
14. 18
15. 7z
16. 35b3c
17.
18.
19.
20.
35 + 6a
15a
12 – 2x 3x x2
15x8
9 + 2m
24m3
21.
22.
23.
24.
25.
26.
27.
28.
29. a b.
c. about 0.8 h
189 – 9n
7n3 x2
20x2
17m – 47
(m + 2)(m – 7)
a2 + 9a + 12
(a + 3)(a + 5)
a2 + 12a + 15
4(a + 3)
c2 + 7c + 20
(c + 5)(c + 3)
4t 2 + 5t + 5
t 2(t + 1)
18a + 3
(2a + 1)(2a – 1) r
0.7r 17 7r
12-6

84. Adding and Subtracting Rational Expressions
ALGEBRA 1 LESSON 12-6
30.
31.
32.
33.
34.
35.
36.
37.
38.
39.
–y 2 + 2y + 2
3y + 1
h 2 + h + 1
2t 2 – 7
r – 2k – 6
9 + p3
–3 – x – z
xy 2z
k + 3km
2m2
12c – 15a
abc
c 3 – a3
10x + 15
x + 2
–21t + 33
2t – 3 6x
(x – 3)(x + 3)2
40.
41. The student added the
terms in the denominators.
42. a ;
b ;
c. Yes; they both represent
the time it takes to make
a round trip.
43. Answers may vary. Sample:
Not always; the numerator
may contain a factor of the LCD.
44. Answers may vary.
Sample: , ;
k – 1
k – 6 2 r
1.25r 18 5r d
0.8d 9 2d 2w
w + 3
3w 2
w – 3
3w3 + 11w2 – 6w
(w + 3)(w – 3)
45.
46. 8
47.
48.
49.
50.
51. –
52.
53.
54.
55. D
8x 2 – 1 x
– 3x – 5
x(x – 5)
32x
x – 5 4x 1
2x(x – 5)
d + 3
d + 4
–x3 + 6x2 + 35x – 50
(a + 12)(a – 5)
x + 4
5a – 8
(a + 2)(a – 5)
12-6

85. Adding and Subtracting Rational Expressions
ALGEBRA 1 LESSON 12-6
56. H
57. D
58. [2] a = total time;
= = =
b = =
= =
The ride took 1 hours.
[1] one computational error
OR no work shown 10 r
r + 3
10(r + 3) + 10r
r(r + 3)
10r r
20r + 30
10(2r + 3)
10(2(12) + 3)
12(12 + 3)
10(24 + 3)
12(15)
10(27)
180
270 3 2
x – 1
60. 5b2 – 10b + 30 –
61. 6
62. 3
63. no solution
64. ± 6.9
65. ± 3.9
66. no solution x2 60
b + 2 1
12-6

86. Adding and Subtracting Rational Expressions
ALGEBRA 1 LESSON 12-6
1. Add + . 2 5x 1 9
2. Add + . 2
3x + 4 6
x + 2
18 + 5x
45x
4(5x + 7)
(3x + 4)(x + 2)
3. Subtract – . 8
x2 – 4 3
4. Subtract – .
6y – 7
y + 2
2y – 3 5
x2 – 4
4(y – 1)
y + 2
5. Add 8 + .
x – 4
x + 3
9x + 20
x + 3
12-6

87. Solving Rational Equations
ALGEBRA 1 LESSON 12-7
(For help, go to Lessons 4-1 and 12-6.)
Solve each proportion. 1 x 3 5 = 1. t 2 2. m 27 3.
Find the LCD of each group of expressions. 4. 3 4n 1 2 n ; 5. 1 3x 2 5 4 ; 6. 1 8y
y 2 5 6 ;
12-7

88. Solving Rational Equations
ALGEBRA 1 LESSON 12-7
Solutions 1. 1 x 3 5 =
3x = 5
x =
or 1 2 2. 3 t 5 2 =
5t = 6
t = 6
or 1 1 3. m 3 27 =
m2 = 81
m = ± = ± 9
4. 4n = 2 • 2 • n; 2 = 2; n = n; LCD = 2 • 2 • n = 4n
5. 3x = 3 • x; 5 = 5; 3x = 3 • x; LCD = 3 • 5 • x = 15x
6. 8y = 2 • 2 • 2 • y; y2 = y • y; 6 = 2 • 3; LCD = 2 • 2 • 2 • 3 • y • y = 24y2
12-7

89. Solving Rational Equations
ALGEBRA 1 LESSON 12-7
Solve + = . 3 8 4 x 14 2x 3 8 4 x 14 2x
+ =
The denominators are 8, x, and 2x. The LCD is 8x.
8x = 8x 3 8 4 x + 14 2x
Multiply each side by the LCD.
8x = 8x 3 8 4 x + 14 2x 1 8x
Use the Distributive Property.
No rational expressions.
Now you can solve.
3x + 32 = 56
3x = 24
Subtract 32 from each side.
x = 8
Divide each side by 3, then simplify.
12-7

90. Solving Rational Equations
ALGEBRA 1 LESSON 12-7
(continued)
Check: 3 8 4 14
2(8) + 7 8 14 16 7 8 =
12-7

91. Solving Rational Equations
ALGEBRA 1 LESSON 12-7 6 x2 5 x
Solve = – 1. Check the solution. 5 x
– 1 6 x2 =
Multiply each side by the LCD, x2. 5 x
– x2 (1) 6 x2 = 1
Use the Distributive Property.
6 = 5x – x2
Simplify.
x2 – 5x + 6 = 0
Collect like terms on one side.
(x – 3)(x – 2) = 0 Factor the quadratic expression.
x – 3 = 0 or x – 2 = 0 Use the Zero-Product Property.
x = 3 or x = 2 Solve.
12-7

92. Solving Rational Equations
ALGEBRA 1 LESSON 12-7
(continued) 6 32 5 3
– 1 6 22 5 2
– 1
Check: 2 3 = 3 2 =
12-7

93. Solving Rational Equations
ALGEBRA 1 LESSON 12-7
Renee can mow the lawn in 20 minutes. Joanne can do the same job in 30 minutes. How long will it take them if they work together?
Define: Let n = the time to complete the job if they work together (in minutes).
Person Work Rate Time Worked Part of
(part of job/min.) (min) Job Done
Renee n
Joanne n 1 20 30 n
Relate: Renee’s part done + Joanne’s part done = complete job.
12-7

94. Solving Rational Equations
ALGEBRA 1 LESSON 12-7
(continued)
Relate: Renee’s part done + Joanne’s part done = complete job. n 20 30 +
Write: = 1 60
= 60(1)
Multiply each side by the LCD, 60.
3n + 2n = 60 Use the Distributive Property.
5n = 60 Simplify.
n = 12 Simplify.
It will take two of them 12 minutes to mow the lawn working together. 1 20
Check:  Renee will do • = of the job, and Joanne will do of the job. Together, they will do = 1, or the whole job. 12 3 5 2 30 • = +
12-7

95. Solving Rational Equations
ALGEBRA 1 LESSON 12-7 4 x2 1
x + 8
Solve = Check the solution. 4 x2 1
x + 8 =
4(x + 8) = x2(1) Write cross products.
4x + 32 = x2 Use the Distributive Property.
x2 – 4x – 32 = 0 Collect terms on one side.
(x – 8)(x + 4) = 0 Factor the quadratic expression.
x – 8 = 0  or  x + 4 = 0 Use the Zero-Product Property.
x = 8  or x = –4 Solve.
12-7

96. Solving Rational Equations
ALGEBRA 1 LESSON 12-7
(continued)
Check: 4 x2 1
x + 8 = 4 82 1
8 + 8
(–4)2
–4 + 8 4 16 1 = 64
12-7

97. Solving Rational Equations
ALGEBRA 1 LESSON 12-7
x + 3
x – 1 4
x – 1
Solve = .
x + 3
x – 1 4 =
(x + 3) (x – 1) = 4(x – 1) Write the cross products.
x2 – x + 3x – 3 = 4x – 4 Use the Distributive Property.
x2 + 2x – 3 = 4x – 4 Combine like terms.
x2 – 2x + 1 = 0 Subtract 4x – 4 from each side.
(x – 1) (x – 1) = 0 Factor.
x – 1 = 0 Use the Zero-Product Property.
x = 1 Simplify.
12-7

98. Solving Rational Equations
ALGEBRA 1 LESSON 12-7
(continued)
Check:
1 + 3
1 – 1 4 4 =
Undefined! There is no division by 0.
The equation has no solution because 1 makes a denominator equal 0.
12-7

99. Solving Rational Equations
ALGEBRA 1 LESSON 12-7
pages 675–677  Exercises
1. –2
2. 3
3. –1
4. 6, –1
5. –
6. –2, 4
7. 1, 4
8. 5
9. 1, 3
10.
11. 1 3 16
12. –4
13. –2
14. –1
15. – h
min
18. 3
19. 10, –10
20. – , 4
21. 4
22. no solution
23. 6 2 3 5 7
24. –14
, 2
26. 3
27. –5, 2
28. –1
29.
30. – , –1
31. 0, 2 h
33. a. 32
b. Answers may vary. Sample: Cross-multiplying; I think it's quicker. 1 2 6 5
12-7

100. Solving Rational Equations
ALGEBRA 1 LESSON 12-7
35. (continued)
c. Yes; the x-values are solutions
to the original equation since
both sides are equal. Ω
37. 20Ω Ω
39. 20Ω
40. Answers may vary.
Sample: =
mi/h
42. 9
43. 0,
44. –1
33 (continued)
c. No; it only works for rational equations that are proportions.
34. a = 4, b = , c = 11, d = –
35. a. y1 = , y2 =
b. (–9.53, 1.07), (–4.16, 1.35),
(–1.12, 5.76), (0.81, 10.16) 1 3 6 x2
(x + 7)2 7 27 2b
b + 2 6b
4b + 3 1 2
12-7

101. Solving Rational Equations
ALGEBRA 1 LESSON 12-7
45. 1
min
47. a. 0.80s
b. 50 – s
c. 0.30(50 – s)
d. 0.8s + 0.3(50 – s) = (0.62)(50)
e. 32
f. 32 L of 80% solution
and 18 L of 30% solution h
49. B
50. G
51. C
52. [2] = ;
2(3 + x) = 16 + x
6 + 2x = 16 + x
x = 10 =
[1] appropriate method,
but with one computational error
53. –
54.
55.
3 + x
16 + x 1 2
3 + 10 13 26 3
x 2y 2z
3h 2 + 2ht + 4h
2(t – 2)(t + 2)
–4k – 61
(k – 4)(k + 10)
12-7

102. Solving Rational Equations
ALGEBRA 1 LESSON 12-7
56.
57.
58.
59.
60.
61.
62. –18, –5
63. 8, 11
64. –23, 1
65. –8, 6
66. –13, –4
67. –91, –1
12-7

103. Solving Rational Equations
ALGEBRA 1 LESSON 12-7
1. Solve = .
2. Solve = .
3. Solve =
4. Solve = .
5. Juanita can wash the car in 30 minutes. Gabe can wash the car in 40 minutes. Working together, how long will it take? x
4x + 3 2
x + 1 5 1 3 –2
3(2 + x) 2x 4
x + 3 –2 1 –4
1, 3
17 min 1 7
12-7

104. Counting Methods and Permutations
ALGEBRA 1 LESSON 12-8
(For help, go to Lessons 12-8 and 4-6.)
You roll a number cube. Find each probability.
1. P(even number) 2. P(prime number)
3. P(a number greater than 5) 4. P(a negative number)
You roll a blue number cube and a yellow number cube. Find each probability.
5. P(blue 1 and yellow 2) 6. P(blue even and yellow odd)
12-8

105. Counting Methods and Permutations
ALGEBRA 1 LESSON 12-8
Solutions
1. P(even number) = P(2, 4, or 6) = =
2. P(prime number) = P(2, 3, or 5) = =
3. P(a number greater than 5) = P(6) =
4. P(a negative number) = = 0
5. P(blue 1 and yellow 2) = P(blue 1) • P(yellow 2) =
6. P(blue even and yellow odd) = P(blue even) • P(yellow odd) = 3 6 1 2 3 6 1 2 1 6 6 1 6 = 36 • 3 6 • 9 36 = 1 4
12-8

106. Counting Methods and Permutations
ALGEBRA 1 LESSON 12-8
Suppose you have three shirts and three ties that coordinate. Make a tree diagram to find the number of possible outfits you have.
Shirts
Ties
Outfits
Shirt 1
Tie1
Tie 2
Tie 3
Shirt 1, Tie 1
Shirt 1, Tie 2
Shirt 1, Tie 3
Shirt 2
Tie1
Tie 2
Tie 3
Shirt 2, Tie 1
Shirt 2, Tie 2
Shirt 2, Tie 3
Shirt 3
Tie1
Tie 2
Tie 3
Shirt 3, Tie 1
Shirt 3, Tie 2
Shirt 3, Tie 3
There are nine possible outfits.
12-8

107. Counting Methods and Permutations
ALGEBRA 1 LESSON 12-8
Suppose there are two routes you can drive to get from Austin, Texas to Dallas, Texas, and four routes from Dallas, Texas to Tulsa, Oklahoma. How many possible routes are there from Austin to Tulsa through Dallas?
2 • 4 = 8
Routes from Austin to Tulsa through Dallas
Routes from Austin to Dallas
Routes from Dallas to Tulsa
There are eight possible routes from Austin to Tulsa.
12-8

108. Counting Methods and Permutations
ALGEBRA 1 LESSON 12-8
In how many ways can 11 students enter a classroom?
There are 11 choices for the first student, 10 for the second, 9 for the third, and so on.
11 • 10 • 9 • 8 • 7 • 6 • 5 • 4 • 3 • 2 • 1 = 39,916,800 Use a calculator.
There are 39,916,800 possible ways in which the students can enter the classroom.
12-8

109. Counting Methods and Permutations
ALGEBRA 1 LESSON 12-8
Simplify 8P5.
Method 1: Use pencil and paper,
8P5 = 8 • 7 • 6 • 5 • 4 The first factor is 8 and there are 5 factors.
= 6720 Simplify.
Method 2: Use a graphing calculator.
Enter the first factor,8.
Use MATH to select nPR in the PRB screen.
Input 5, since there are 5 factors. Press enter.
8P5 = 6720
12-8

110. Counting Methods and Permutations
ALGEBRA 1 LESSON 12-8
Suppose you use five different letters from the 26 letters of the alphabet to make a password. Find the number of possible five-letter passwords.
There are 26 letters in the alphabet. You are finding the number of permutations of 26 letter arranged 5 at a time.
8P5 = 26 • 25 • 24 • 23 • 22 Use a calculator.
= 7,893,600
There are 7,893,600 five-letter passwords in which letters do not repeat.
12-8

111. Counting Methods and Permutations
ALGEBRA 1 LESSON 12-8
pages 682–685  Exercises
1. 10 choices
Shirt 1 Tie 1 S1, T1
Tie 2 S1, T2
Tie 3 S1, T3
Tie 4 S1, T4
Tie 5 S1, T5
Shirt 2 Tie 1 S2, T1
Tie 2 S2, T2
Tie 3 S2, T3
Tie 4 S2, T4
Tie 5 S2, T5
2. 12 menus
Main
Salad   Soup   Course   Menu
C SVC
V B SVB
S S SVS
C SCC
C B SCB
S SCS
C CVC
V B CVB
C S CVS
C CCC
C B CCB
S CCS
12-8

112. Counting Methods and Permutations
ALGEBRA 1 LESSON 12-8
3. a. 8, 10, 10, 10
b. 8,000,000 telephone numbers
4. a. 6
b. 12
5. 3,628,800 orders
arrangements
9. 360
14. 42
16. 12,144
17. 8P6
18. 9P7
19. 8P4
20. a. 2; 2
b. 4
c. No; number of consonants •
number of vowels = number of vowels •
number of consonants.
21. a. 24 b. 1 24
12-8

113. Counting Methods and Permutations
ALGEBRA 1 LESSON 12-8
22. a. 24 b.
c. Answers may vary. Sample:
No; if someone tries to guess
your password, they’ll probably
try your name or initials first.
23. 3
24. 2
25. 5
26. a. 10,000
b. With repetition; there are more
permutations when repetition
is allowed. 1
14,950
27. a. 260,000 license plates
b. 23,920,000 license plates
28. a. Check students’ work.
b. Check students’ work.
29. a. 17,576 codes
b. 17,526 codes
30. a. 35,152 call letters
b. 913,952 call letters
31. a. 18,278 companies
b. 12,338,352 companies
c. NASDAQ; 12,320,074
more companies
32. a. 2
b. 6
c. (n – 1)!
12-8

114. Counting Methods and Permutations
ALGEBRA 1 LESSON 12-8
41. 3
42. –8, 6
43. –10, 1
44. –8, 5
, –5
46. AB = 7, AC = 5
47. BC 34, AC 54
48. AB 21, BC 5
49. AB 48, AC 7
50. –6 ±
51. no solution
52. –4 ±
53. – ±
54. –2 ± 5
55. 5, 13
33. 7
34. a. 60 numbers
b. 6 numbers c. b
35. False; if a = 3 and b = 2,
then (a – b)! = 1! = 1,
but a! – b! = 3! – 2! = 4.
36. 72
37. 16
38. 4
39. 24
40. 7 2 97 2 1 10 9 10 1 2 3 10
12-8

115. Counting Methods and Permutations
ALGEBRA 1 LESSON 12-8
1. Jeff has five shirts, three pairs of pants, and two ties. How many possible outfits can he make?
2. Three of 18 students are to be selected for the student government positions of president, vice-president, and treasurer. In how many different ways can the three positions be filled?
3. Calculate 6P3.
4. An Italian restaurant offers four different choices of pasta, three choices of sauce, and two choices of meat. How many different dishes can you order that each consist of one pasta choice, one sauce, and one meat?
30 outfits
4896 ways
120 24
12-8

116. Evaluate each expression. 1. 5P3 2. 6P3 3. 7P3 4. 7P4
Combinations
ALGEBRA 1 LESSON 12-9
(For help, go to Lessons 12-8 and 4-6.)
Evaluate each expression.
1. 5P3 2. 6P3 3. 7P3 4. 7P4
A and B are independent events. Find P(A and B) for the given probabilities.
5. P(A) = , P(B) = 6. P(A) = , P(B) =
7. P(A) = , P(B) = 8. P(A) = 0.35, P(B) = 0.2 1 3 3 4 1 8 5 9 9 10 5 6
12-9

117. 8. P(A and B) = P(A) • P(B) = (0.35)(0.2) = 0.07 • =
Combinations
ALGEBRA 1 LESSON 12-9
(For help, go to Lessons 12-8 and 4-6.)
Solutions
1. 5P3 = 5 • 4 • 3 = P3 = 6 • 5 • 4 = 120
3. 7P3 = 7 • 6 • 5 = P4 = 7 • 6 • 5 • 4 = 840
5. P(A and B) = P(A) • P(B) =
6. P(A and B) = P(A) • P(B) =
7. P(A and B) = P(A) • P(B) =
8. P(A and B) = P(A) • P(B) = (0.35)(0.2) = 0.07 1 3 4 • =
1 • 3
3 • 4 1 8 5 9 • = 72 5 6 • = 9 10 3 4
3 • 3 • 5
5 • 2 • 3 • 2
2 • 2
12-9

118. Write using permutation notation.
Combinations
ALGEBRA 1 LESSON 12-9
Simplify 9C6.
9C6 =
9P6
6P6
Write using permutation notation.
Write the product represented by the notation.
9 • 8 • 7 • 6 • 5 • 4
6 • 5 • 4 • 3 • 2 • 1 =
Simplify.
= 84
12-9

119. Method 1: Use pencil and paper.
Combinations
ALGEBRA 1 LESSON 12-9
Eighteen people enter a talent contest. Awards will be given to the top ten finishers. How many different groups of ten winners can be chosen?
The order in which the top ten winners are listed once they are chosen does not distinguish one group of winners from another. You need the number of combinations of 18 potential winners chosen 10 at a time. Evaluate 18C10.
Method 1: Use pencil and paper.
18C10 =
18P10
10P10
Write using permutation notation.
= 43,758
Use a calculator. =
18 • 17 • 16 • 15 • 14 • 13 • 12 • 11 • 10 • 9
10 • 9 • 8 • 7 • 6 • 5 • 4 • 3 • 2 • 1
12-9

120. Method 2: Use a graphing calculator.
Combinations
ALGEBRA 1 LESSON 12-9
(continued)
Method 2: Use a graphing calculator.
Enter the first factor, 18.
Use to select nCr in the PRB screen.
Input 10, since there are ten factors. Press .
18C10 = 43,758.
There are 43,758 different ten-person groups of winners that can be chosen from a group of 18 people.
12-9

121. Three of the eight CDs are country.
Combinations
ALGEBRA 1 LESSON 12-9
Suppose you have eight new CDs (three rock, two jazz, and three country). If you choose the CDs at random, what is the probability that the first one is country and the second one is jazz?
Three of the eight CDs are country.
probability the first choice is country = 3 8
Two of the remaining CDs are jazz.
probability the second choice is jazz = 2 7
P(country, then jazz) = 3 8 2 7 •
Multiply the probabilities. 6 56 =
Multiply. = 3 28
Simplify.
The probability that you choose country, then jazz is 3 28
12-9

122. There are 12 pens in all. Eight of the pens are red.
Combinations
ALGEBRA 1 LESSON 12-9
Suppose you have eight red pens and four black pens in a box. You choose five pens without looking. What is the probability that all the pens you choose are red?
There are 12 pens in all. Eight of the pens are red.
number of favorable outcomes = 8C5
number of ways to choose 5 pens from the 8 red pens
number of possible outcomes = 12C5
number of ways to choose 5 red pens from 12 possible pens
12-9

123. Use the definition of probability. P(5 red pens) =
Combinations
ALGEBRA 1 LESSON 12-9
(continued)
Use the definition of probability.
P(5 red pens) =
total number of outcomes
number of favorable outcomes
8C5
12C5 =
Substitute. = 56
792
Simplify each expression. 7 99
Simplify.
The probability that you will chose five red pens is , or about 7%. 7 99
12-9

124. 20. combination, since order is not important
ALGEBRA 1 LESSON 12-9
pages 689–691  Exercises
1. 1
2. 6
3. 15
4. 20
5. 15
6. 6
7. 28
8. 28
9. 21
10. 21
12. 56
13.
14. a. 45
b. 3 c. d.
15. a. 56
b. 1
16. 10
17. 1
18. 35
19. 4
20. combination, since order
is not important
21. permutation, since order
is important
22. a.
b. 6
c. 6
d. 45 2 15 1 56 5 28
5040
12-9

125. 23. a. Answers may vary. Sample: It is a combination problem
Combinations
ALGEBRA 1 LESSON 12-9
23. a. Answers may vary. Sample:
It is a combination problem
because order is not important.
b. 45
c. Yes; Each line segment joins
two points and each handshake
connects two people.
24. a. 59,280 sequences
b sequences c.
d. Answers may vary. Sample:
It is unlikely someone will guess
the right sequence with more
than 59,000 possibilities. 1 40
25. Answers may vary. Sample:
Both permutations and combinations
are arrangements of some or all of a
group of objects. However, permutations
take into account order, and
combinations do not.
26. 4
27. 8
28. a. 24 b.
29. Check students’ work.
30. a. 792
b. 36 4 22
12-9

126. b. Answers may vary. Sample: It is the function ƒ(x) = ,
Combinations
ALGEBRA 1 LESSON 12-9
31. always
32. sometimes
33. always
34.
35. a. 15
b. 6; 3
c. 20
d. 300
36. a.
b. Answers may vary. Sample:
It is the function ƒ(x) = , 1 28
x(x – 1) 2
36. (continued)
which uses the combination
formula xC2 = for x.
This represents the number of
combination groups of 2.
c. Groups can only be made from sets of objects, which means they must be integers.
37. B
38. F
39. C
40. [2] There are 10 possible pizzas;
combination; order does
not matter; x!
2!(x – 2)!
12-9

127. [1] incorrect explanation OR minor error 41. 12 42. 5040 43. 840 44. 9
Combinations
ALGEBRA 1 LESSON 12-9
49.
50. 5
51. –2
52. 0, 16
53. –6.81, 0.81
, 6.70
55. –5.46, 1.46
56. –1.24, 1.35
57. –6, 3
58. –0.05, 13.38 1 10
40. [2] (continued)
5C2 = = = 10
[1] incorrect explanation OR
minor error
41. 12
44. 9
45. 23
46. 92,610,000 license plates
47. 2
48. no solution
5P2
5 • 4
2 • 1 1 2
12-9

128. Combinations 1. Simplify 30C4.
ALGEBRA 1 LESSON 12-9
1. Simplify 30C4.
2. You have just received a box of chocolates. There are three turtles, four caramels, and three chocolate-covered cherries. If you choose two chocolates at random, what is the probability that the first one is caramel and the second is a turtle?
3. There are twelve congressmen who want to be on the same committee. Eight are Republicans and four are Democrats. If three people are chosen for the committee, what is the probability that three Republicans are chosen?
4. A friend has loaned you five books and you plan to read all of them. In how many ways could you read the five books?
27,405 2 15 14 55
120 ways
12-9

129. Rational Expressions and Functions
ALGEBRA 1 CHAPTER 12
1. 30
2. 7.8
3. –24
5. D h
7. x = 0; y = 0
8. x = 0; y = 0
9. x = 0; y = 3
10. x = 0; y = 3
12-A

130. Rational Expressions and Functions
ALGEBRA 1 CHAPTER 12
11. Answers may vary. Sample:
In direct variation and in inverse
variation, the variables are related
to each other by a constant. But in
direct variation, that number is the
ratio of any corresponding pair of
input and output values. Distance
traveled varies directly with average
speed. In inverse variation, that
number is the product of any
corresponding pair of input and
output values. The cost per person
of splitting a $14 pizza varies inversely
with the number of people who
are sharing it.
12.
13.
14.
15.
16. Answers may vary. Sample:
17. 4x + 3 –
18. x3 – 2x2 + 4x – 8
19. 2x3 – 2x2 – x +
20. 2x2 – 5x – 2 + h
22. –6, –2
23. 4
x + 2 4 1
6x2 14
3w – 5
2c(3c – 1)
3(c + 5)(c – 3)
(x – 6)(x – 3) 10 3x 7
2x – 1 17
3x + 2 2
12-A

131. Rational Expressions and Functions
ALGEBRA 1 CHAPTER 12
24. no solution
25.
26.
27.
28.
29. Permutation; order is important;
870 different pairs.
30. Combination; order is not important;
15 different ways.
31. Combination; order is not important;
15 different pairs of toppings.
32. 15
33. 70
t 2 + 5t + 5
t(t + 1)
n + 9
n(n + 1) 1
y + 3
7b 2 + 6b – 4
3b(b + 2)
34. 5
35. 35
36. 4
37. 20,160
,800
39. 10
different combinations
41. 33 91
12-A